Question

In quadrilateral abcd,seg ad parallel seg bc.diagonal ac and diagonal bd intersect each other in point p.then show that ap/pd=pc/bp?

Answer 1

First we have to make construction..
Draw seg ST such that it is parallel to seg BC & S-P-T.
And by using property of three || lines and thier transversals we get,
AS/SB=DT/TC.......(1)
Now,by using basic proportionality theorem in triangles ABC & DCB we get,
AS/SB=AP/PC.......(2)
DT/TC=PD/BP...........(3)
From (1),(2) & (3),
AP/PC=PD/BP
i.e. AP/PD=PC/BP

Answer 2

Answer:

draw Seg ST such that S-P-T

by bpt,

AS/SB=DT/TC

AS/SB=AP/PC

DP/PB=DT/TC

therefore,

AS/SB=DT/TC=AP/PC=DP/PB

hence

AP/PD=PC=BP

proved